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Joined: Dec 2000
Posts: 4,289
Member
You can buy a heck of a lot of paper and pencils for $599.
You can still know, and not guess. [Linked Image]

Joined: May 2005
Posts: 30
J
Member
Bottom line -- #10 for 3.6% voltage drop at last light.

Since I'm a student, I got out my trusty Ugly's book. Using the Ohm's Law wheel on the front I calculated that a 100w bulb on a 120v circuit has a resistance of 144 ohms. [E*E/W = 120*120/100]. Call the bulb resistance B. Since I'm a student, I'll ignore the fact that resistance of a light bulb varies with the applied voltage.

Next I looked up the AC resistance of of #14 through #6 wire. It is given for 1000ft so I divided it by 20 to get the resistance for 50ft:

14 - .16; 12 - .10; 10 - .06; 8 - .04; 6 - .02. Call the wire resistance C.

Now picture the circuit as a ladder with each rung a light. Label each light as R300 through R309, with R300 at the source and R309 farthest away. Label the left rail that connects R300 with R301 as R101. Number the rest of the left rails R102 through R109. Number the corresponding right rails as R201 through R209.

Now the fun begins: using the series and parallel resistance formulas calculate the circuit's total resistance.

First do the R109, R309, R209 group which is simply 2 * C + B, which for #10 is 144.12 ohms.

To calculate the resistance of the R108, R308 (in parallel with the R109, R309, R209 group), and R208 you need to first calculate R308 and the parallel circuit which is 72.03 [1/((1/Ra)+(1/Ry)) 1/((1/R308)+(1/144.12))] call this Rp and then add 2 * C which is 72.15 [R108+Rp+R208 2 * .06 + 72.03].

You repeat this calculation back to R101, R301, R201 which gives you a table of resistance values at the source side of the R101, R201 terminals to the R109, R209 terminals:

resistors resistance
Rxx1 16.42
Rxx2 18.38
Rxx3 20.91
Rxx4 24.30
Rxx5 29.06
Rxx6 36.22
Rxx7 48.19
Rxx8 72.15
Rxx9 144.12

Now that we know the resistances we can calculate the current, starting at the source. The source voltage is 120v, the resistance is 16.42 ohms, so the current is 7.31a [E/R = I 120/16.42]. The we can calculate the voltage drop in R101 and R201 to be .88v [R*I = E 2*C * 16.42] This leaves 119.12v across R301. Now we repeat the calculations and get the following table:

resistors current total volts drop
Rxx1 7.31a 120v .88v
Rxx2 6.48a 119.12v .78v
Rxx3 5.66a 118.35v .68v
Rxx4 4.84a 117.67v .58v
Rxx5 4.03a 117.09v .48v
Rxx6 3.22a 116.60v .39v
Rxx7 2.41a 116.22v .29v
Rxx8 1.61a 115.93v .19v
Rxx9 .80a 115.73v .10v

This gives a total voltage drop of 4.36v or 3.64%


JFW
Joined: Sep 2005
Posts: 33
N
Member
WOW what answer. but you can only have a 6 volt drop across a 120volt circuitfrom the feeders to the last lampost. and a
3.6 volt drop from the last overcurrent device/disconect.

Joined: Mar 2005
Posts: 38
D
Member
Quote

You can buy a heck of a lot of paper and pencils for $599.
LOL, I think you missed the point of cost vs price. Unless you're retired, your labor cost is far more than a piece of paper and a couple of pencils and that's where the ROI lays with Volts. Also, the download is free for 10-days so....how much was that piece of paper and pencil again?

And currently Volts sells for $499.00 from Dolphins Software and only $449.00 from ECN's store.

Quote

Doing it this way is more fun and keeps us sharp. I don't want to "dumb it down" by relying on software.
Agreed. But if you're going to attempt these types of computations I suggest that you use the normal engineering methods established by IEEE Standard 141 in place of formulae guessing and assumations. The Standard 141 is featured in IEEE's Red Book which can be purchased on EBay. Also, this type of problem requires a Sequential Voltage Drop method of computation, not Ohm's Law.

Additionally, please don't forget to account for ambient and terminal temperatures when selecting your conductor size along with a continuous load increase factor of 125%.

And with an ambient temperature of 105ºF, terminal temperature at 90ºC, metal conduit, type THHN insulation with copper conductors, 125% conductor capacity, power factor at unity; the first conductor size is #8 with 0.43%VD and the last is #10 with 2.87% VD per IEEE Standard 141 Exact Voltage Drop Formulae. Computation time was about 25 seconds 'cause I type slow.

[This message has been edited by DiverDan (edited 10-30-2005).]


Dolphins Software
Joined: Feb 2002
Posts: 840
C
Member
ngoody24,

Limiting voltage drop is not an NEC requirement. VD is mentioned in an FPN which is not enforceable.

Now practically speaking, we have many reasons to limit voltage drop as much as possible, but there is no code requirement to limit it.

Peter


Peter
Joined: Dec 2004
Posts: 161
G
Member
Quote
you could also use loop or "Ring" (not in the over seas sense) circuit

Roger,

Rings work well when wiring around a space like a room, but if the lights are all in a line you will only be doubling your wire cost.

(Just had to rise to that one! [Linked Image])

DiverDan, how could you calculate voltage drops without resorting to Ohms Law at some stage? Is the Sequential Voltage Drop method not based on Ohms Law?

Joined: Mar 2005
Posts: 38
D
Member
Quote

Limiting voltage drop is not an NEC requirement. VD is mentioned in an FPN which is not enforceable.

True the NEC does not enforce VD...but the manufactures of the electrical devices do have limitations of operation, performance and efficiency concerning VD.

I'm adding this section because I think gideonr posted as I was writting this reply.

About Voltage Drop and Ohm's Law. Ohm's law works great with DC systems but falls short on AC systems with power factors and alternating current sinusoid waves.

[This message has been edited by DiverDan (edited 10-30-2005).]


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